Pascal's Triangle for Python

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轮回少年
轮回少年 2020-11-30 06:13

As a learning experience for Python, I am trying to code my own version of Pascal\'s triangle. It took me a few hours (as I am just starting), but I came out with this code:

11条回答
  •  情深已故
    2020-11-30 06:30

    OK code review:

    import math
    
    # pascals_tri_formula = [] # don't collect in a global variable.
    
    def combination(n, r): # correct calculation of combinations, n choose k
        return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))
    
    def for_test(x, y): # don't see where this is being used...
        for y in range(x):
            return combination(x, y)
    
    def pascals_triangle(rows):
        result = [] # need something to collect our results in
        # count = 0 # avoidable! better to use a for loop, 
        # while count <= rows: # can avoid initializing and incrementing 
        for count in range(rows): # start at 0, up to but not including rows number.
            # this is really where you went wrong:
            row = [] # need a row element to collect the row in
            for element in range(count + 1): 
                # putting this in a list doesn't do anything.
                # [pascals_tri_formula.append(combination(count, element))]
                row.append(combination(count, element))
            result.append(row)
            # count += 1 # avoidable
        return result
    
    # now we can print a result:
    for row in pascals_triangle(3):
        print(row)
    

    prints:

    [1]
    [1, 1]
    [1, 2, 1]
    

    Explanation of Pascal's triangle:

    This is the formula for "n choose k" (i.e. how many different ways (disregarding order), from an ordered list of n items, can we choose k items):

    from math import factorial
    
    def combination(n, k): 
        """n choose k, returns int"""
        return int((factorial(n)) / ((factorial(k)) * factorial(n - k)))
    

    A commenter asked if this is related to itertools.combinations - indeed it is. "n choose k" can be calculated by taking the length of a list of elements from combinations:

    from itertools import combinations
    
    def pascals_triangle_cell(n, k):
        """n choose k, returns int"""
        result = len(list(combinations(range(n), k)))
        # our result is equal to that returned by the other combination calculation:
        assert result == combination(n, k)
        return result
    

    Let's see this demonstrated:

    from pprint import pprint
    
    ptc = pascals_triangle_cell
    
    >>> pprint([[ptc(0, 0),], 
                [ptc(1, 0), ptc(1, 1)], 
                [ptc(2, 0), ptc(2, 1), ptc(2, 2)],
                [ptc(3, 0), ptc(3, 1), ptc(3, 2), ptc(3, 3)],
                [ptc(4, 0), ptc(4, 1), ptc(4, 2), ptc(4, 3), ptc(4, 4)]],
               width = 20)
    [[1],
     [1, 1],
     [1, 2, 1],
     [1, 3, 3, 1],
     [1, 4, 6, 4, 1]]
    

    We can avoid repeating ourselves with a nested list comprehension:

    def pascals_triangle(rows):
        return [[ptc(row, k) for k in range(row + 1)] for row in range(rows)]
    
    >>> pprint(pascals_triangle(15))
    [[1],
     [1, 1],
     [1, 2, 1],
     [1, 3, 3, 1],
     [1, 4, 6, 4, 1],
     [1, 5, 10, 10, 5, 1],
     [1, 6, 15, 20, 15, 6, 1],
     [1, 7, 21, 35, 35, 21, 7, 1],
     [1, 8, 28, 56, 70, 56, 28, 8, 1],
     [1, 9, 36, 84, 126, 126, 84, 36, 9, 1],
     [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1],
     [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1],
     [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1],
     [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1],
     [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]
    

    Recursively defined:

    We can define this recursively (a less efficient, but perhaps more mathematically elegant definition) using the relationships illustrated by the triangle:

     def choose(n, k): # note no dependencies on any of the prior code
         if k in (0, n):
             return 1
         return choose(n-1, k-1) + choose(n-1, k)
    

    And for fun, you can see each row take progressively longer to execute, because each row has to recompute nearly each element from the prior row twice each time:

    for row in range(40):
        for k in range(row + 1):
            # flush is a Python 3 only argument, you can leave it out,
            # but it lets us see each element print as it finishes calculating
            print(choose(row, k), end=' ', flush=True) 
        print()
    
    
    1
    1 1
    1 2 1
    1 3 3 1
    1 4 6 4 1
    1 5 10 10 5 1
    1 6 15 20 15 6 1
    1 7 21 35 35 21 7 1
    1 8 28 56 70 56 28 8 1
    1 9 36 84 126 126 84 36 9 1
    1 10 45 120 210 252 210 120 45 10 1
    1 11 55 165 330 462 462 330 165 55 11 1
    1 12 66 220 495 792 924 792 495 220 66 12 1
    1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
    1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
    1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
    1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
    1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
    1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 ...
    

    Ctrl-C to quit when you get tired of watching it, it gets very slow very fast...

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